# Thread: Summation of Finite Series using Standard Results

1. ## Summation of Finite Series using Standard Results

Sorry, this one is a little long-winded, but I hope it's clear! I've come across a problem that I'm struggling to solve because it *requires* the use of certain standard results as the starting point. The problem is to find the solution of

$\sum\limits_{r = 1}^n {\frac{1}{r\left( {r + 1} \right)}}$

I can solve this problem quite easily by considering the form of each term in the sum, or through the use of partial fractions (where I find an ultimate answer of $\frac{n}{n + 1}$, but I'm supposed to solve it using one (or more) of the following standard results (I'll include all of them, but I imagine only the final one is relevant):

$\sum\limits_{r = 1}^n {r} = \frac{1}{2} {n ( {n + 1} )}$

$\sum\limits_{r = 1}^n {r^2} = \frac{1}{6}$n(n + 1)(2n + 1)

$\sum\limits_{r = 1}^n {r^3} = \frac{1}{4}$n2(n + 1)2

$\sum\limits_{r = 1}^n {r\left( {r + 1} \right) } = \frac{1}{3}$ n(n + 1)(n + 2)

I can't figure out how to get from the right-hand side of the standard result to the final answer of $\frac{n}{n + 1}$. I assume I'm missing something really simple about the relationship between $\sum{f(x)}$ and $\sum{\frac{1}{f(x)}}$, but I just can't see it. Any help would be greatly appreciated. Thanks very much.

2. ## Re: Summation of Finite Series using Standard Results

Originally Posted by britmath
Sorry, this one is a little long-winded, but I hope it's clear! I've come across a problem that I'm struggling to solve because it *requires* the use of certain standard results as the starting point. The problem is to find the solution of

$\sum\limits_{r = 1}^n {\frac{1}{r\left( {r + 1} \right)}}$
$\sum\limits_{r = 1}^n {\frac{1}{{r\left( {r + 1} \right)}}} = \sum\limits_{r = 1}^n {\left( {\frac{1}{r} - \frac{1}{{r + 1}}} \right)} = 1 - \frac{1}{{n + 1}}$

3. ## Re: Summation of Finite Series using Standard Results

Thanks for the response Plato. Unfortunately, though the partial fraction decomposition makes complete sense, the problem expects the starting point to be one of the four standard results mentioned in the original post, so I'm expecting the solution to look something like this:

$\sum\limits_{r = 1}^n {r\left( {r + 1} \right) } = \frac{1}{3}$ n(n + 1)(n + 2)

therefore

$\sum\limits_{r = 1}^n {\frac{1}{r\left( {r + 1} \right)}}$ = some manipulation of the equation above

I feel like I'm missing something really simple to get started on this problem, but I can't figure out what that is! Thanks again.